Cho tan(a + b) = 3, tan(a – b) = 2. Tính: tan2a, tan2b.

Ta có:

tan2a = tan[(a + b) + (a – b)]

         $= \frac{{\tan \left( {a + b} \right) + \tan \left( {a - b} \right)}}{{1 - \tan \left( {a + b} \right)\tan \left( {a - b} \right)}} = \frac{{3 + 2}}{{1 - 3.2}} = \frac{5}{{ - 5}} = - 1$;

tan2b = tan[(a + b) ‒ (a – b)]

          $= \frac{{\tan \left( {a + b} \right) - \tan \left( {a - b} \right)}}{{1 + \tan \left( {a + b} \right)\tan \left( {a - b} \right)}} = \frac{{3 - 2}}{{1 + 3.2}} = \frac{1}{7}$.